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midterm10sols

midterm10sols - L Vandenberghe EE103 Midterm solutions...

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Unformatted text preview: L. Vandenberghe 10/26/10 EE103 Midterm solutions Problem 1 (10 points). A square matrix A is called normal if AA T = A T A . Show that if A is normal and nonsingular, then the matrix Q = A- 1 A T is orthogonal. Solution. We show that Q T Q = I : Q T Q = ( A- 1 A T ) T ( A- 1 A T ) = AA- T A- 1 A T = A ( AA T )- 1 A T = A ( A T A )- 1 A T = AA- 1 A- T A T = I. On line 2 we use ( BC ) T = C T B T . On lines 3 and 5 we use the property that ( BC )- 1 = C- 1 B- 1 if B and C are invertible. On line 4 we use the definition of normal matrix. Problem 2 (10 points). Let a = ( a 1 , a 2 , a 3 ), b = ( b 1 , b 2 , b 3 ), c = ( c 1 , c 2 , c 3 ), d = ( d 1 , d 2 , d 3 ) be four given points in R 3 . The points do not lie in one plane. Algebraically, this can be expressed by saying that the vectors b − a , c − a , d − a are linearly independent, i.e. , y 1 ( b − a )+ y 2 ( c − a )+ y 3 ( d − a ) = 0 only if y 1 = y 2 = y 3 = 0. Suppose we are given the distances of a point x = ( x 1 , x 2 , x 3 ) to the four points: bardbl x − a bardbl = r a , bardbl x − b bardbl = r b , bardbl x − c bardbl = r c , bardbl x − d bardbl = r d . Write a set of linear equations Ax = f , with A nonsingular, from which the coordinates x 1 , x 2 , x 3 can be computed. Explain why the matrix A is nonsingular. Solution. Squaring the norms gives four nonlinear equations bardbl x bardbl 2 − 2 a T x + bardbl a bardbl 2 = r 2 a bardbl x bardbl 2 − 2 b T x + bardbl b bardbl 2 = r 2 b bardbl x bardbl 2 − 2 c T x + bardbl c bardbl 2 = r 2 c bardbl x bardbl 2 − 2 d T x + bardbl d bardbl 2 = r 2 d . Subtracting the first equation from the three others gives 2( a − b ) T x = r 2 b − r 2 a + bardbl a bardbl 2 − bardbl b bardbl 2 2( a − c ) T x = r 2 c − r 2 a + bardbl a bardbl 2 − bardbl c bardbl 2 2( a − d...
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midterm10sols - L Vandenberghe EE103 Midterm solutions...

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